06 Stress Strain

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    Stress strain relations Prof Schierle 1

    Stre

    ss

    Strain

    Stress/ StrainRelations

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    Stress strain relations Prof Schierle 2

    Stress / strain test

    = /

    Pull bar to cause stress and strain

    Record load P

    Compute stress f

    f = P/A

    A = cross section area

    Record strainL

    Compute unit strain = L / L L = unstressed length

    Plot measure points on stress strain graph Draw line through plotted points

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    Stress / Strain RelationsHookes Law for isotropicmaterial

    (material of equal properties in any direction)

    Stress causes strain deformation

    Stress/strain relations are visualized

    by a spring as substitute for a rod

    to amplify stress / strain relations

    1 Elongation under tension

    2 Shortening under compression3 Stress / strain graph

    P Applied load

    L Unstressed length

    L Elongation/shortening under load Unit strain =L/L

    f Stress f = P/A (A = cross section area)

    E Elastic modulus E = f /

    (defines stiffness of material)

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    Stress / Strain

    1 Test load (increasing from 1 to 5 k)

    2 Stress-strain graph

    For each load the strain is recorded

    A line through recorded points

    defines stress / strain relations

    Slope defines Elastic Modulus E

    E = f /

    f = P / A = stress

    P = applied load

    A = test bar cross-section area

    = L/L = unit strainL = elongation / shortening

    L = unstressed length

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    5 Elastic material (rubber)

    Resumes initial length after load

    6 Plastic material (clay)

    Remains deformed after load

    S = permanent set

    7 Brittle material (concrete)

    8 Ductile material (steel)

    Stress / strain graphs

    1 Test loads (1-5)

    2 Stress / strain graph

    3 Linear material

    Strain increase linear with stress4 Nonlinear material

    Strain increase nonlinear with stress

    Stress strain relations Copyright Prof Schierle 2014 5

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    1 Poissons ratio (shrinks / expands material

    normal to applied load)

    2 Creep = deformation over time (critical in

    concrete)C = Creep

    T = Time

    3 Elastic/plastic material4 Steel (idealized graph)

    5 Steel (mild steel)

    6 High strength steel

    7 Concrete

    8 Wood (under tensile stress)

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    Stress / Strain

    3 Elastic / plastic material

    E Elastic range

    P Plastic range

    Slope of decreasing stress,

    parallel to increasing stress,causes permanent deformation

    4 Steel graph (idealized)

    A Proportional limit

    B Elastic limitC Yield point

    CD Yield plateau

    E Ultimate strength

    F Breaking point

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    Mild steel (Fy = 36 ksi yield stress)(Mild steel is increasingly replaced by

    high-strength steel of 50 ksi yield strength)

    Allowable stress for Fy = 36 ksi

    Type of stress Allowable stresses

    Axial stress Fa = 21.6 ksi (0.6 Fy)Bending stress Fb = 21.6 ksi (0.6 Fy)

    Shear stress Fv = 14.4 ksi (0.4 Fy)

    Allowable stress for Fy = 50 ksiType of stress Allowable stresses

    Axial stress Fa = 30 ksi (0.6 Fy)

    Bending stress Fb = 30 ksi (0.6 Fy)

    Shear stress Fv = 20 ksi (0.4 Fy)

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    1,5001.5Masonry

    3,0003Concrete

    30,00030Steel

    1,4001.4Wood

    Stiffness E (ksi)

    Elastic modulus

    Strength F (ksi)Material

    Typical strength F vs. stiffness E

    Axial deflection LDerivation of formula

    f =P/A =L / L

    E = f/

    E =(P/A) / (L/L)

    E = P L / (AL)

    L = PL / AE

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    Examples

    Elevator cables

    Assume

    4 cables each, 60% metallic area

    Breaking strength Fy = 210 ksi

    Allowable stress (210 ksi / {3x4}) Fa = 17.5 ksiElastic Modulus E=16,000 ksi

    Cable length each L = 700

    Load P = 8 k

    Metallic areaAm = 4 x .6 r2= 4 x .6 0.252 Am = 0.47 in2

    Stress

    f = P / A = 8 / 0.47 f = 17 ksi

    17 < 17.5, ok

    Elongation

    L = PL / AE = 8k x 700x12 / (0.47x16000) L = 8.9

    ~

    70%metallic

    ~

    60%metallic

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    Cable elevator

    Suspended on 4 to 8 wire ropeseach alone strong enough to

    support the elevator

    Safety breaks block elevator

    if all cables break

    Air cushion slows free-fallin case of failure

    Shock absorber cushionsimpact in case of failure

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    Settlement effect on curtain wall

    Assume

    Steel structure, aluminum curtain wall

    2-story mullion, length L = 30x12 L = 360

    Column stress increase during construction f = 15 ksi

    Elastic modulus (steel) E = 29,000 ksi

    Column shortening due to stress

    L = PL / AE

    Since f = P/ A

    L = f L/E = 15 ksi x 360 / 29000 L = 0.19

    Note:To allow settlement without stress in mullion,

    provide minimum mullion expansion joints

    (also verify joint width for thermal expansion)

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    Suspended structure settlementAssume:

    Steel columns, strand hangers

    10 stories @ 14 = 10x14x12 L = 1680Average column stress f = 18 ksi

    Average strand stress f = 60 ksi

    Elastic modulus (steel) E = 29,000 ksi

    Elastic modulus (strand) E = 22,000 ksi

    Column strain

    L = PL/AE

    Since f = P/A L = f L/EL = 18 ksi x 1680/29000 L = 1.0

    Strand strain

    L = 60ksi x 1680 / 22000 L = 4.6

    Differential settlement L = 5.6Note:

    To reduce differential settlement:

    Prestress strands to reduceL by half and adjust

    floor levels for DL and partial LL

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    a p p y e n d

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